Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $16.4$ years; the standard deviation is $1.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $12.6$ years.
$16.4$ $14.5$ $18.3$ $12.6$ $20.2$ $10.7$ $22.1$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $16.4$ years. We know the standard deviation is $1.9$ years, so one standard deviation below the mean is $14.5$ years and one standard deviation above the mean is $18.3$ years. Two standard deviations below the mean is $12.6$ years and two standard deviations above the mean is $20.2$ years. Three standard deviations below the mean is $10.7$ years and three standard deviations above the mean is $22.1$ years. We are interested in the probability of a gorilla living less than $12.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $12.6$ years and the other half $({2.5\%})$ will live longer than $20.2$ years. The probability of a particular gorilla living less than $12.6$ years is ${2.5\%}$.